second order ode $u''(x)=k^2u(x)$, $u(\pi)=0$.

Question

Would appreciate if someone would solve explicitly the following second order ode:

$u''(x)=k^2u(x)$

$u(\pi)=0$.

I know the result is supposed to be $u(x)=c_k\sinh(k(x-\pi))$

Thank you.

Answer 1

Substitute $$u(x)=e^{\lambda x}$$ then you will get $$\lambda^2-k^2=0$$ so we get $$(k-\lambda)(k+\lambda)=0$$ Can you finish?

Answer 2

$$u''(x)=k^2u(x) \to R^2-k^2=0 \to R=\mp k$$ The general solution is $$u(x)=K_1e^{kx}+K_2e^{-kx}$$ Since $u(\pi)=0$ $$K_1e^{k\pi}+K_2e^{-k\pi}=0 \to K_2=-K_1e^{2k\pi}$$ Therefore : $$u=K_1(e^{kx}-e^{2k\pi-kx})$$ $$u=\underbrace{2K_1e^{k\pi}}_{\text {constante K}}\left (\frac {e^{k(x-\pi)}-e^{-k(x-\pi)}} 2 \right)$$ $$\boxed{u(x)=K\sinh(k(x-\pi))}$$